Pertemuan 13 - 14CLOSED CONDUIT FLOW 1

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Reynolds Experiment• Reynolds Number

• Laminar flow: Fluid moves in smooth streamlines• Turbulent flow: Violent mixing, fluid velocity at a point

varies randomly with time• Transition to turbulence in a 2 in. pipe is at V=2 ft/s,

so most pipe flows are turbulent

2lowfTurbulent4000

flowTransition40002000

flowLaminar2000

Re

Vh

VhVD

f

f

Laminar Turbulent

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Shear Stress in Pipes• Steady, uniform flow in a pipe: momentum flux is zero

and pressure distribution across pipe is hydrostatic, equilibrium exists between pressure, gravity and shear forces

D

Lhhh

ds

dhD

zp

ds

dD

sDds

dzsAsA

ds

dp

sDWAsds

dpppAF

f

s

021

0

0

0

0

44

)]([4

)(0

)(sin)(0

• Since h is constant across the cross-section of the pipe (hydrostatic), and –dh/ds>0, then the shear stress will be zero at the center (r = 0) and increase linearly to a maximum at the wall.

• Head loss is due to the shear stress.

• Applicable to either laminar or turbulent flow• Now we need a relationship for the shear stress in

terms of the Re and pipe roughness

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Darcy-Weisbach Equation

)(Re,

)(Re,

;Re;

,,:variablesRepeating

),(

),,,,(

20

20

20

321

214

0

D

eFV

D

eF

V

VD

e

DV

F

eDVF

V D e

ML-1T-2 ML-3 LT-1 ML-1T-1 L L

)(Re,82

)(Re,82

)(Re,4

4

2

2

2

0

D

eFf

g

V

D

Lfh

D

eF

g

V

D

L

D

eFV

D

L

D

Lh

f

f

Darcy-Weisbach Eq. Friction factor

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Laminar Flow in Pipes• Laminar flow -- Newton’s law of viscosity is valid:

2

0

20

20

2

14

44

2

2

2

r

r

ds

dhrV

ds

dhrCC

ds

dhrV

drds

dhrdV

ds

dhr

dr

dV

dr

dV

dy

dV

ds

dhr

dy

dV

• Velocity distribution in a pipe (laminar flow) is parabolic with maximum at center.

2

0max 1

r

rVV

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Discharge in Laminar Flow

ds

dhD

ds

dhrQ

rr

ds

dh

rdrrrds

dhVdAQ

rrds

dhV

r

r

128

8

2

)(

4

)2()(4

)(4

4

40

0

220

2

022

0

220

0

0

ds

dhDV

A

QV

32

2

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Head Loss in Laminar Flow

2

21

12212

2

2

2

32

)(32

32

32

32

D

VLh

hhh

ssD

Vhh

dsD

Vdh

DV

ds

dh

ds

dhDV

f

f

Re

64

2

2/)(Re

64

2/))((64

2/

2/32

32

2

2

2

2

2

2

2

fV

D

Lfh

VD

L

VD

L

DV

V

V

D

VL

D

VLh

f

f

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Nikuradse’s Experiments

• In general, friction factor

– Function of Re and roughness

• Laminar region

– Independent of roughness

• Turbulent region– Smooth pipe curve

• All curves coincide @ ~Re=2300

– Rough pipe zone• All rough pipe curves flatten

out and become independent of Re

Re

64f

Blausius

Re 4/1k

f Rough

Smooth

Laminar Transition Turbulent

Blausius OK for smooth pipe

)(Re,D

eFf

Re

64f

2

9.010Re

74.5

7.3log

25.0

D

ef

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Moody Diagram

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Pipe Entrance• Developing flow

– Includes boundary layer and core, – viscous effects grow inward from the wall

• Fully developed flow– Shape of velocity profile is same at all points

along pipe

flowTurbulent 4.4Re

flowLaminar Re06.01/6D

Le

eL

Entrance length LeFully developed

flow region

Region of linear pressure drop

Entrance pressure drop

Pressure

x

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Entrance Loss in a Pipe• In addition to frictional losses, there are minor

losses due to – Entrances or exits– Expansions or contractions– Bends, elbows, tees, and other fittings– Valves

• Losses generally determined by experiment and then corellated with pipe flow characteristics

• Loss coefficients are generally given as the ratio of head loss to velocity head

• K – loss coefficent– K ~ 0.1 for well-rounded inlet (high Re)– K ~ 1.0 abrupt pipe outlet– K ~ 0.5 abrupt pipe inlet

Abrupt inlet, K ~ 0.5

g

VKh

g

V

hK L

L

2or

2

2

2

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Elbow Loss in a Pipe• A piping system may have many minor losses which

are all correlated to V2/2g • Sum them up to a total system loss for pipes of the

same diameter

• Where,

m

mm

mfL KD

Lf

g

Vhhh

2

2

lossheadTotalLh

lossheadFrictionalfh

mhm fittingfor lossheadMinor

mKm fittingfor tcoefficienlossheadMinor

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EGL & HGL for Losses in a Pipe• Entrances, bends, and other flow transitions cause the

EGL to drop an amount equal to the head loss produced by the transition.

• EGL is steeper at entrance than it is downstream of there where the slope is equal the frictional head loss in the pipe.

• The HGL also drops sharply downstream of an entrance

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Ex(10.2)Given: Liquid in pipe has = 8 kN/m3. Acceleration = 0. D =

1 cm, = 3x10-3 N-m/s2. Find: Is fluid stationary, moving up, or moving down? What

is the mean velocity?Solution: Energy eq. from z = 0 to z = 10 m

smV*x*

).*(.V

μL

γDhV

γD

μLVh

mh

h

h

zp

g

Vhz

p

g

V

-

L

L

L

L

L

L

/04.11010332

0108000251

32

32

upward) (moving25.1

108

90

108000

000,110

8000

000,200

22

3

2

2

2

22

22

211

21

1

1

2

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Ex (10.4)• Given: Oil (S = 0.97, m = 10-2 lbf-s/ft2) in 2 in pipe, Q =

0.25 cfs.• Find: Pressure drop per 100 ft of horizontal pipe.• Solution:

ftpsi/100791122

46111002103232

(laminar)36010

)12/2(*46.11*94/1*97.0Re

/46.114/)12/2(

25.0

22

2

2

.)/(

.**-*

D

μLVΔp

VD

sftA

QV

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Ex. (10.8)Given: Kerosene (S=0.94, =0.048 N-s/m2). Horizontal 5-cm

pipe. Q=2x10-3 m3/s.Find: Pressure drop per 10 m of pipe.Solution:

cfs

sftV

VV

VVg

VγD

μLV

g

α

g

Vα

γD

μLV

γD

μLVh

zγ

p

g

Vαhz

γ

p

g

Vα

L

L

32

5

22

2

522

222

2

22

22

2

22

22

211

21

1

10*23.14/(0.25/12)**1.6A*VQ

(laminar)129310*4

)12/25.0(*6.1*94.1*8.0Re

/60.1

01.1645.8

05.0)32/1(*4.62*8.0

10*10*4*32

2

2

05.032

2

002

325.000

32

22

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Ex. (10.34)

2/62.0,/300,12 mNsmN

Given: Glycerin@ 20oC flows commercial steel pipe. Find: hSolution:

mγD

μLVhh

VDVD

hzγ

pz

γ

ph

zγ

phz

γ

p

zγ

p

g

Vαhz

γ

p

g

Vα

L

L

L

L

42.2)02.0(*300,12

)6.0)(1)(62.0(3232

(laminar)5.2310*1.5

02.0*6.0Re

)(

22

22

4

22

11

22

11

22

22

211

21

1

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Ex. (10.43)Given: FigureFind: Estimate the elevation required in the upper reservoir to

produce a water discharge of 10 cfs in the system. What is the minimum pressure in the pipeline and what is the pressure there?

Solution:

ftz

sftA

QV

D

LfKKK

g

V

D

LfKKKh

zhz

zγ

p

g

Vαhz

γ

p

g

Vα

Ebe

EbeL

L

L

1332.32*2

73.1275.100.14.0*25.0100

/73.121*4/

10

75.101

430*025.0;0.1;(assumed)4.0;5.0

22

0000

22

2

1

2

221

22

22

211

21

1

55

2

22

1

2

1

2

11

21

1

10*910*14.1

1*73.12Re

59.0)53.1(*4.62

35.1

2.32*2

73.12

1

300025.04.05.00.17.110133

22

2*100

22

VD

psigp

ft

g

V

D

LfKK

g

Vzz

γ

p

zγ

p

g

Vhz

zγ

p

g

Vαhz

γ

p

g

Vα

b

beb

bb

bbb

L

bbb

bL

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Ex. (10.68)

ftxksftx s425 105.1;/1022.1

ftDD

g

DQ

D

g

V

D

Lfh f

06.8

984,331

2

))4//((1000*015.01

2

5

22

2

Given: Commercial steel pipe to carry 300 cfs of water at 60oF with a head loss of 1 ft per 1000 ft of pipe. Assume pipe sizes are available in even sizes when the diameters are expressed in inches (i.e., 10 in, 12 in, etc.).

Find: Diameter.Solution:

Assume f = 0.015

Relative roughness: 00002.006.8

105.1 4

x

D

ks

Get better estimate of f

65

2

109.31022.1)06.8)(4/(

300Re

)4/()4/(

Re

xx

D

QD

D

Q

VD

f=0.010

.8943.7

656,221

5

inftDD

Use a 90 in pipe

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Ex. (10.81)

mh

h

zγ

p

g

Vαhz

γ

p

g

Vα

g

Q

h

fLD

g

DQ

D

Lf

g

V

D

Lfh

f

f

L

f

f

45.14

109810

000,60

9810

000,300

22

8

2

))4//((

2

22

22

211

21

1

5/1

2

2

222

Given: The pressure at a water main is 300 kPa gage. What size pipe is needed to carry water from the main at a rate of 0.025 m3/s to a factory that is 140 m from the main? Assume galvanized-steel pipe is to be used and that the pressure required at the factory is 60 kPa gage at a point 10 m above the main connection.

Find: Size of pipe.Solution:

Assume f = 0.020

m

g

Q

h

fLD

f

100.081.9

)025.0(140

45.14

02.08

8

5/1

2

2

5/1

2

2

Relative roughness:

022.0

0015.0100

15.0

f

D

ks

Friction factor:

mD 102.0020.0

022.0100.0

5/1

Use 12 cm pipe

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Ex. (10.83)Given: The 10-cm galvanized-steel pipe is 1000 m long and

discharges water into the atmosphere. The pipeline has an open globe valve and 4 threaded elbows; h1=3 m and h2 = 15 m.

Find: What is the discharge, and what is the pressure at A, the midpoint of the line?

Solution:

002

)41(1200

222

22

22

211

21

1

g

V

D

LfKKK

zγ

p

g

Vαhz

γ

p

g

Vα

bve

L

D = 10-cm and assume f = 0.025

46

32

2

2

1071031.1

1.0*942.0Re

/0074.0)10.0)(4/(942.0

/942.01.265

24

)1.0

1000025.09.0*4105.01(24

xx

VD

smVAQ

smV

gV

Vg

So f = 0.025

kPap

mgγ

p

g

V

D

LfK

γ

p

zγ

p

g

Vαhz

γ

p

g

Vα

A

A

bA

LAAA

A

8.90)26.9(*9810

6.9152

)942.0()

1.0

500025.09.0*2(

2)2(15

22

2

2

22

22

2

2

Near cavitation pressure, not good!

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Ex. (10.95)

Given: If the deluge through the system shown is 2 cfs, what horsepower is the pump supplying to the water? The 4 bends have a radius of 12 in and the 6-in pipe is smooth.

Find: HorsepowerSolution:

55

22

2

22

22

22

211

21

1

1017.41022.1

)2/1(*18.10Re

611.12

/18.10)2/1)(4/(

2

)45.01(2

6003000

22

xx

VD

ftg

V

sftA

QV

D

LfK

g

Vh

hzγ

p

g

Vαhz

γ

p

g

Vα

bp

Lp

hphQ

p

ft

h

p

p

4.24550

6.107

))2/1(

17000135.019.0*45.01(611.13060

So f = 0.0135